Respuesta :
This is an incomplete question, here is a complete question.
Consider a buffer solution that is 0.50 M in NH₃ and 0.20 M in NH₄Cl. For ammonia, pKb = 4.75.
Calculate the pH of 1.0 L of the original buffer upon addition of 0.010 mol of solid NaOH to the original buffer solution.
Answer : The pH of solution is, 9.68
Explanation : Given,
[tex]pK_b=4.75[/tex]
Concentration of NH₃ (base) = 0.50 M
Concentration of NH₄Cl (salt) = 0.20 M
Concentration of NaOH = [tex]\frac{Moles}{Volume}=\frac{0.010mol}{1.0L}=0.01M[/tex]
After addition of NaOH:
Concentration of NH₃ (base) = 0.50 + 0.01 = 0.51 M
Concentration of NH₄Cl (salt) = 0.20 - 0.01 = 0.19 M
First we have to calculate the value of pOH.
Using Henderson Hesselbach equation :
[tex]pOH=pK_b+\log \frac{[Salt]}{[Base]}[/tex]
Now put all the given values in this expression, we get:
[tex]pOH=4.75+\log (\frac{0.19}{0.51})[/tex]
[tex]pOH=4.32[/tex]
Now we have to calculate the pH.
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.32\\\\pH=9.68[/tex]
Therefore, the pH of solution is, 9.68
The pH of 1.0 L of the buffer solution is 9.68.
What is the pH of a solution?
The pH of a solution is the degree of acidity or alkalinity of the solution. It is usually represented on a pH scale ranging from 0 to 14.
From the given information, Let's assume that; a buffer solution contains:
- 0.50 M in NH₃ and,
- 0.20 M in NH₄Cl
where:
- The pkb = 4.75
Firstly, we need to determine the concentration of the Solid NaOH in the solution.
[tex]\mathbf{Concentration \ of \ NaOH = \dfrac{moles }{volume}}[/tex]
[tex]\mathbf{Concentration \ of \ NaOH = \dfrac{0.010 \ mol }{1.0 \ L }}[/tex]
Concentration of NaoH = 0.01 M
Now, at the time when NaOH is added to the base(NH3) and the salt(NH4Cl), their concentration becomes:
- NH₃ = 0.50 + 0.01 = 0.51 M
- NH₄Cl = 0.20 - 0.01 = 0.19 M
Thus, the next process is to determine the value of pOH by using the Henderson-Hasselbach equation:
[tex]\mathbf{pOH = pKb + log \dfrac{[salt]}{[base]} }[/tex]
[tex]\mathbf{pOH = pKb + log \dfrac{[NH_4Cl]}{[NH_3]} }[/tex]
[tex]\mathbf{pOH = 4.75 + log \Big( \dfrac{0.19}{0.51} \Big)}[/tex]
pOH = 4.32
Therefore, the pH is finally calculated as:
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 4.32
pH = 9.68
Learn more about the pH of a solution here:
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